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Since $\exp(\cdot)$ is local lipschitz, the following sde has s strong solution $$ \mathrm{d}X_s=\exp(X_s)\mathrm{d}B_s, X(0)=1, $$ where $B$ is a Brownian Motion. I wander if the following expression holds? $$\mathbb{E}\int_{0}^T\exp(2X(s))<\infty.$$

yxyt
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  • Yes, it admits a local-in-time unique strong solution since $x \mapsto \exp(x)$ is locally Lipschitz, i.e up to some random explosion time $e\left(\omega, X(0)\right)$ there exists a unique strong solution to your equation. For details, refer to Ikeda & Watanabe's Stochastic Differential Equations and Diffusion Processes (particularly theorems 2.3 and 3.1 of chapter 4). – Shiva May 21 '23 at 11:35
  • Thanks for the reference, and i also wander if $E \int _0^T \exp(2X(s))\mathrm{d}s<\infty$? – yxyt May 21 '23 at 14:06
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    For $T$ large enough no. Set $u(t,x) = \mathbb{E}\left[\exp(2X(t))|X(0) = x\right]$, then a computation using the generator shows that $$\partial_t u(t, x) = \mathbb{E}\left[2\exp(4X(t))|X(0) = x\right] \geq 2\mathbb{E}\left[\exp(2X(t))|X(0) = x\right]^2 = 2(u(t,x))^2.$$ We can lower bound $u$ by $\phi$ the solution of the following equation $$\partial_t \phi(t,x) = 2\left(\phi(t,x)\right)^2 \quad \phi(0, x) = u(0, x).$$ Solving this equation shows that for fixed $x$, $\phi(t,x) \to \infty$ as $t \to \frac{1}{2\phi(0, x)}$. Thus if $T\geq \exp(-2X(0))/2$ the integral expression is not finite. – Shiva May 21 '23 at 15:53
  • I get that, thanks! – yxyt May 22 '23 at 02:08

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